Questions: 1. (1,0 ponto) Faça o balanceamento das seguintes equações químicas abaixo: a) Al(s) + Cl2(a) -> AlCl3(s) b) C2H4(g) + O2(g) -> CO2(g) + H2O(g) 2. (1,0 ponto) Cite pelo menos três características que os elementos metálicos possuem. 3. (1,0 ponto) Considere a seguinte reação: 2 Mg(s) + O2(g) -> 2 MgO(s). Se você quer preparar 1 g de MgO, qual massa de magnésio é preciso usar? a) 0,60 g de Mg b) 0,30 g de Mg c) 0,20 g de Mg d) 0.40 g de Mg e) Nenhuma das alternativas 4. (1,0 ponto) Qual é a massa molar do carbonato de cálcio, CaCO3? a) 150,09 g / mol b) 100,09 g / mol c) 164,09 g / mol d) 68.09 g / mol e) Nenhuma das alternativas

Solution Steps

To balance the equation, we need to ensure the number of atoms of each element is the same on both sides: \[ 2 \mathrm{Al}_{(s)} + 3 \mathrm{Cl}_{2(a)} \rightarrow 2 \mathrm{AlCl}_{3(s)} \]

To balance the equation, we need to ensure the number of atoms of each element is the same on both sides: \[ \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) + 3 \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)} + 2 \mathrm{H}_{2} \mathrm{O}_{(g)} \]

\[ \boxed{2 \mathrm{Al}_{(s)} + 3 \mathrm{Cl}_{2(a)} \rightarrow 2 \mathrm{AlCl}_{3(s)}} \] \[ \boxed{\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) + 3 \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)} + 2 \mathrm{H}_{2} \mathrm{O}_{(g)}} \]

1. High electrical conductivity
2. High thermal conductivity
3. Malleability and ductility

\[ \boxed{\text{1. High electrical conductivity, 2. High thermal conductivity, 3. Malleability and ductility}} \]

Given the reaction: \[ 2 \mathrm{Mg}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s}) \]

The molar mass of Mg is 24.305 g/mol, and the molar mass of MgO is 40.304 g/mol. To find the mass of Mg needed to produce 1 g of MgO, we use stoichiometry:

\[ \text{Moles of MgO} = \frac{1 \text{ g}}{40.304 \text{ g/mol}} = 0.0248 \text{ mol} \]

Since the reaction shows a 1:1 molar ratio between Mg and MgO: \[ \text{Moles of Mg} = 0.0248 \text{ mol} \]

\[ \text{Mass of Mg} = 0.0248 \text{ mol} \times 24.305 \text{ g/mol} = 0.6028 \text{ g} \]

The closest answer is: \[ \boxed{0.60 \text{ g de Mg}} \]

Final Answer